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PHXI05:LAWS OF MOTION

363095 In the setup shown, a 200 $N$ block is supported in equilibrium with the help of a string and a spring. Extension in the spring is 4 $c m$ . Force constant of the spring is closest to $[g = 10 m / s^{2}]$
supporting img

1

30 N / m

2

2500 N / m

3

3000 N / m

4

4000 N / m

Explanation:

Let $T_{1}$ be the tension in the spring and $T_{2}$ be the tension in the right string
in equilibrium. Acc to Lami’s theorem
$\frac{200}{\sin (90^{\circ})} = \frac{T_{1}}{\sin (90 + 53)}$
$\frac{200}{1} = \frac{T_{1}}{\cos 53}$
$T_{1} = 120 N = K x \Rightarrow K = 3000 N / m$
supporting img

PHXI05:LAWS OF MOTION

363096 Two masses of $10 k g$ and $20 k g$ are connected by a massless spring, as shown in the figure. A force of $100 N$ acts on the $20 k g$ mass at the instant when the $10 k g$ mass has an acceleration of $6 m / s^{2}$ towards right, the acceleration of the $20 k g$ mass is
supporting img

1

5 m / s^{2}

2

2 m / s^{2}

3

8 m / s^{2}

4

6 m / s^{2}

Explanation:

Let $a$ be the acceleration of block of mass $20 k g$ .
supporting img
Let $T$ be the tensions in the spring
$T = 10 \times 6 = 60 N$
$100 - 60 = 20 (1) \Rightarrow a = 2 m / s^{2}$

PHXI05:LAWS OF MOTION

363097 Figure shows a ball of mass $m$ connected with two ideal springs of force constant $k$ , kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration (in $m / s^{2}$ ) of ball?
supporting img

1

7 m / s^{2}

2

5 m / s^{2}

3

10 m / s^{2}

4

3 m / s^{2}

Explanation:

Let initial extension in spring is $x_{0}$ , then
$2 k x_{0} \cos 45^{\circ} = m g \sin 45^{\circ} \Rightarrow 2 k x_{0} = m g$
Let after cutting right spring, according of ball is $A_{1}$ along the inclined plane and $a_{2}$ perpendicular to $a_{1}$
$\Rightarrow \frac{m g}{\sqrt{2}} - \frac{m g}{2} \frac{1}{\sqrt{2}} = m a_{1}$
$\Rightarrow a_{1} = \frac{g}{2 \sqrt{2}}$
Then
$m g \sin 45^{\circ} \Rightarrow k x_{0} \cos 45^{\circ} = m a_{1}$
$k x_{0} \sin 45^{\circ} = m a_{2} \Rightarrow a_{2} = \frac{g}{2 \sqrt{2}}$
Net acceleration $a = \sqrt{a_{1}^{2} + a_{2}^{2}} = \frac{g}{2} = 5 m / s^{2}$

PHXI05:LAWS OF MOTION

363098 A block of mass 3 $k g$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figure. If spring $A$ is cut from lower point at $t = 0$ then, find acceleration of block in $m s^{- 2}$ at $t = 0$ . (ignore rotational effect.)
supporting img

1

5

2

10

3

15

4

0

Explanation:

Before cutting the spring
$2 k x = 3 g \Rightarrow k x = 15$
After cutting
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363099 Two masses $10 k g$ and $20 k g$ respectively are connected by a massless spring as shown in figure. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown is figure, the $10 k g$ mass has acceleration of $12 m / s^{2} .$ The value of acceleration of $20 k g$ mass is
supporting img

1

4 m / s^{2}

2

10 m / s^{2}

3

20 m / s^{2}

4

30 m / s^{2}

Explanation:

Equation of motion of mass $m_{1} = 10 k g$ is
$F_{1} = m_{1} a_{1} = 10 \times 12 = 120 N$
Force on $10 k g$ mass is $120 N$ to the right. As
action and reaction are equal and opposite, the
reactions force $F$ - on $20 k g$ mass $F = 120 N$ to
the left.
$∴$ Equation of motion of mass
$m_{2} = 20 k g$ is
$200 - F = 20 a_{2}$
$200 - 120 = 20 a_{2}$
$20 a_{2} = 80$
$a_{2} = \frac{80}{20} = 4 m / s^{2}$

PHXI05:LAWS OF MOTION

363095 In the setup shown, a 200 $N$ block is supported in equilibrium with the help of a string and a spring. Extension in the spring is 4 $c m$ . Force constant of the spring is closest to $[g = 10 m / s^{2}]$
supporting img

1

30 N / m

2

2500 N / m

3

3000 N / m

4

4000 N / m

Explanation:

Let $T_{1}$ be the tension in the spring and $T_{2}$ be the tension in the right string
in equilibrium. Acc to Lami’s theorem
$\frac{200}{\sin (90^{\circ})} = \frac{T_{1}}{\sin (90 + 53)}$
$\frac{200}{1} = \frac{T_{1}}{\cos 53}$
$T_{1} = 120 N = K x \Rightarrow K = 3000 N / m$
supporting img

PHXI05:LAWS OF MOTION

363096 Two masses of $10 k g$ and $20 k g$ are connected by a massless spring, as shown in the figure. A force of $100 N$ acts on the $20 k g$ mass at the instant when the $10 k g$ mass has an acceleration of $6 m / s^{2}$ towards right, the acceleration of the $20 k g$ mass is
supporting img

1

5 m / s^{2}

2

2 m / s^{2}

3

8 m / s^{2}

4

6 m / s^{2}

Explanation:

Let $a$ be the acceleration of block of mass $20 k g$ .
supporting img
Let $T$ be the tensions in the spring
$T = 10 \times 6 = 60 N$
$100 - 60 = 20 (1) \Rightarrow a = 2 m / s^{2}$

PHXI05:LAWS OF MOTION

363097 Figure shows a ball of mass $m$ connected with two ideal springs of force constant $k$ , kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration (in $m / s^{2}$ ) of ball?
supporting img

1

7 m / s^{2}

2

5 m / s^{2}

3

10 m / s^{2}

4

3 m / s^{2}

Explanation:

Let initial extension in spring is $x_{0}$ , then
$2 k x_{0} \cos 45^{\circ} = m g \sin 45^{\circ} \Rightarrow 2 k x_{0} = m g$
Let after cutting right spring, according of ball is $A_{1}$ along the inclined plane and $a_{2}$ perpendicular to $a_{1}$
$\Rightarrow \frac{m g}{\sqrt{2}} - \frac{m g}{2} \frac{1}{\sqrt{2}} = m a_{1}$
$\Rightarrow a_{1} = \frac{g}{2 \sqrt{2}}$
Then
$m g \sin 45^{\circ} \Rightarrow k x_{0} \cos 45^{\circ} = m a_{1}$
$k x_{0} \sin 45^{\circ} = m a_{2} \Rightarrow a_{2} = \frac{g}{2 \sqrt{2}}$
Net acceleration $a = \sqrt{a_{1}^{2} + a_{2}^{2}} = \frac{g}{2} = 5 m / s^{2}$

PHXI05:LAWS OF MOTION

363098 A block of mass 3 $k g$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figure. If spring $A$ is cut from lower point at $t = 0$ then, find acceleration of block in $m s^{- 2}$ at $t = 0$ . (ignore rotational effect.)
supporting img

1

5

2

10

3

15

4

0

Explanation:

Before cutting the spring
$2 k x = 3 g \Rightarrow k x = 15$
After cutting
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363099 Two masses $10 k g$ and $20 k g$ respectively are connected by a massless spring as shown in figure. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown is figure, the $10 k g$ mass has acceleration of $12 m / s^{2} .$ The value of acceleration of $20 k g$ mass is
supporting img

1

4 m / s^{2}

2

10 m / s^{2}

3

20 m / s^{2}

4

30 m / s^{2}

Explanation:

Equation of motion of mass $m_{1} = 10 k g$ is
$F_{1} = m_{1} a_{1} = 10 \times 12 = 120 N$
Force on $10 k g$ mass is $120 N$ to the right. As
action and reaction are equal and opposite, the
reactions force $F$ - on $20 k g$ mass $F = 120 N$ to
the left.
$∴$ Equation of motion of mass
$m_{2} = 20 k g$ is
$200 - F = 20 a_{2}$
$200 - 120 = 20 a_{2}$
$20 a_{2} = 80$
$a_{2} = \frac{80}{20} = 4 m / s^{2}$

PHXI05:LAWS OF MOTION

363095 In the setup shown, a 200 $N$ block is supported in equilibrium with the help of a string and a spring. Extension in the spring is 4 $c m$ . Force constant of the spring is closest to $[g = 10 m / s^{2}]$
supporting img

1

30 N / m

2

2500 N / m

3

3000 N / m

4

4000 N / m

Explanation:

Let $T_{1}$ be the tension in the spring and $T_{2}$ be the tension in the right string
in equilibrium. Acc to Lami’s theorem
$\frac{200}{\sin (90^{\circ})} = \frac{T_{1}}{\sin (90 + 53)}$
$\frac{200}{1} = \frac{T_{1}}{\cos 53}$
$T_{1} = 120 N = K x \Rightarrow K = 3000 N / m$
supporting img

PHXI05:LAWS OF MOTION

363096 Two masses of $10 k g$ and $20 k g$ are connected by a massless spring, as shown in the figure. A force of $100 N$ acts on the $20 k g$ mass at the instant when the $10 k g$ mass has an acceleration of $6 m / s^{2}$ towards right, the acceleration of the $20 k g$ mass is
supporting img

1

5 m / s^{2}

2

2 m / s^{2}

3

8 m / s^{2}

4

6 m / s^{2}

Explanation:

Let $a$ be the acceleration of block of mass $20 k g$ .
supporting img
Let $T$ be the tensions in the spring
$T = 10 \times 6 = 60 N$
$100 - 60 = 20 (1) \Rightarrow a = 2 m / s^{2}$

PHXI05:LAWS OF MOTION

363097 Figure shows a ball of mass $m$ connected with two ideal springs of force constant $k$ , kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration (in $m / s^{2}$ ) of ball?
supporting img

1

7 m / s^{2}

2

5 m / s^{2}

3

10 m / s^{2}

4

3 m / s^{2}

Explanation:

Let initial extension in spring is $x_{0}$ , then
$2 k x_{0} \cos 45^{\circ} = m g \sin 45^{\circ} \Rightarrow 2 k x_{0} = m g$
Let after cutting right spring, according of ball is $A_{1}$ along the inclined plane and $a_{2}$ perpendicular to $a_{1}$
$\Rightarrow \frac{m g}{\sqrt{2}} - \frac{m g}{2} \frac{1}{\sqrt{2}} = m a_{1}$
$\Rightarrow a_{1} = \frac{g}{2 \sqrt{2}}$
Then
$m g \sin 45^{\circ} \Rightarrow k x_{0} \cos 45^{\circ} = m a_{1}$
$k x_{0} \sin 45^{\circ} = m a_{2} \Rightarrow a_{2} = \frac{g}{2 \sqrt{2}}$
Net acceleration $a = \sqrt{a_{1}^{2} + a_{2}^{2}} = \frac{g}{2} = 5 m / s^{2}$

PHXI05:LAWS OF MOTION

363098 A block of mass 3 $k g$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figure. If spring $A$ is cut from lower point at $t = 0$ then, find acceleration of block in $m s^{- 2}$ at $t = 0$ . (ignore rotational effect.)
supporting img

1

5

2

10

3

15

4

0

Explanation:

Before cutting the spring
$2 k x = 3 g \Rightarrow k x = 15$
After cutting
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363099 Two masses $10 k g$ and $20 k g$ respectively are connected by a massless spring as shown in figure. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown is figure, the $10 k g$ mass has acceleration of $12 m / s^{2} .$ The value of acceleration of $20 k g$ mass is
supporting img

1

4 m / s^{2}

2

10 m / s^{2}

3

20 m / s^{2}

4

30 m / s^{2}

Explanation:

Equation of motion of mass $m_{1} = 10 k g$ is
$F_{1} = m_{1} a_{1} = 10 \times 12 = 120 N$
Force on $10 k g$ mass is $120 N$ to the right. As
action and reaction are equal and opposite, the
reactions force $F$ - on $20 k g$ mass $F = 120 N$ to
the left.
$∴$ Equation of motion of mass
$m_{2} = 20 k g$ is
$200 - F = 20 a_{2}$
$200 - 120 = 20 a_{2}$
$20 a_{2} = 80$
$a_{2} = \frac{80}{20} = 4 m / s^{2}$

PHXI05:LAWS OF MOTION

363095 In the setup shown, a 200 $N$ block is supported in equilibrium with the help of a string and a spring. Extension in the spring is 4 $c m$ . Force constant of the spring is closest to $[g = 10 m / s^{2}]$
supporting img

1

30 N / m

2

2500 N / m

3

3000 N / m

4

4000 N / m

Explanation:

Let $T_{1}$ be the tension in the spring and $T_{2}$ be the tension in the right string
in equilibrium. Acc to Lami’s theorem
$\frac{200}{\sin (90^{\circ})} = \frac{T_{1}}{\sin (90 + 53)}$
$\frac{200}{1} = \frac{T_{1}}{\cos 53}$
$T_{1} = 120 N = K x \Rightarrow K = 3000 N / m$
supporting img

PHXI05:LAWS OF MOTION

363096 Two masses of $10 k g$ and $20 k g$ are connected by a massless spring, as shown in the figure. A force of $100 N$ acts on the $20 k g$ mass at the instant when the $10 k g$ mass has an acceleration of $6 m / s^{2}$ towards right, the acceleration of the $20 k g$ mass is
supporting img

1

5 m / s^{2}

2

2 m / s^{2}

3

8 m / s^{2}

4

6 m / s^{2}

Explanation:

Let $a$ be the acceleration of block of mass $20 k g$ .
supporting img
Let $T$ be the tensions in the spring
$T = 10 \times 6 = 60 N$
$100 - 60 = 20 (1) \Rightarrow a = 2 m / s^{2}$

PHXI05:LAWS OF MOTION

363097 Figure shows a ball of mass $m$ connected with two ideal springs of force constant $k$ , kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration (in $m / s^{2}$ ) of ball?
supporting img

1

7 m / s^{2}

2

5 m / s^{2}

3

10 m / s^{2}

4

3 m / s^{2}

Explanation:

Let initial extension in spring is $x_{0}$ , then
$2 k x_{0} \cos 45^{\circ} = m g \sin 45^{\circ} \Rightarrow 2 k x_{0} = m g$
Let after cutting right spring, according of ball is $A_{1}$ along the inclined plane and $a_{2}$ perpendicular to $a_{1}$
$\Rightarrow \frac{m g}{\sqrt{2}} - \frac{m g}{2} \frac{1}{\sqrt{2}} = m a_{1}$
$\Rightarrow a_{1} = \frac{g}{2 \sqrt{2}}$
Then
$m g \sin 45^{\circ} \Rightarrow k x_{0} \cos 45^{\circ} = m a_{1}$
$k x_{0} \sin 45^{\circ} = m a_{2} \Rightarrow a_{2} = \frac{g}{2 \sqrt{2}}$
Net acceleration $a = \sqrt{a_{1}^{2} + a_{2}^{2}} = \frac{g}{2} = 5 m / s^{2}$

PHXI05:LAWS OF MOTION

363098 A block of mass 3 $k g$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figure. If spring $A$ is cut from lower point at $t = 0$ then, find acceleration of block in $m s^{- 2}$ at $t = 0$ . (ignore rotational effect.)
supporting img

1

5

2

10

3

15

4

0

Explanation:

Before cutting the spring
$2 k x = 3 g \Rightarrow k x = 15$
After cutting
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363099 Two masses $10 k g$ and $20 k g$ respectively are connected by a massless spring as shown in figure. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown is figure, the $10 k g$ mass has acceleration of $12 m / s^{2} .$ The value of acceleration of $20 k g$ mass is
supporting img

1

4 m / s^{2}

2

10 m / s^{2}

3

20 m / s^{2}

4

30 m / s^{2}

Explanation:

Equation of motion of mass $m_{1} = 10 k g$ is
$F_{1} = m_{1} a_{1} = 10 \times 12 = 120 N$
Force on $10 k g$ mass is $120 N$ to the right. As
action and reaction are equal and opposite, the
reactions force $F$ - on $20 k g$ mass $F = 120 N$ to
the left.
$∴$ Equation of motion of mass
$m_{2} = 20 k g$ is
$200 - F = 20 a_{2}$
$200 - 120 = 20 a_{2}$
$20 a_{2} = 80$
$a_{2} = \frac{80}{20} = 4 m / s^{2}$

PHXI05:LAWS OF MOTION

363095 In the setup shown, a 200 $N$ block is supported in equilibrium with the help of a string and a spring. Extension in the spring is 4 $c m$ . Force constant of the spring is closest to $[g = 10 m / s^{2}]$
supporting img

1

30 N / m

2

2500 N / m

3

3000 N / m

4

4000 N / m

Explanation:

Let $T_{1}$ be the tension in the spring and $T_{2}$ be the tension in the right string
in equilibrium. Acc to Lami’s theorem
$\frac{200}{\sin (90^{\circ})} = \frac{T_{1}}{\sin (90 + 53)}$
$\frac{200}{1} = \frac{T_{1}}{\cos 53}$
$T_{1} = 120 N = K x \Rightarrow K = 3000 N / m$
supporting img

PHXI05:LAWS OF MOTION

363096 Two masses of $10 k g$ and $20 k g$ are connected by a massless spring, as shown in the figure. A force of $100 N$ acts on the $20 k g$ mass at the instant when the $10 k g$ mass has an acceleration of $6 m / s^{2}$ towards right, the acceleration of the $20 k g$ mass is
supporting img

1

5 m / s^{2}

2

2 m / s^{2}

3

8 m / s^{2}

4

6 m / s^{2}

Explanation:

Let $a$ be the acceleration of block of mass $20 k g$ .
supporting img
Let $T$ be the tensions in the spring
$T = 10 \times 6 = 60 N$
$100 - 60 = 20 (1) \Rightarrow a = 2 m / s^{2}$

PHXI05:LAWS OF MOTION

363097 Figure shows a ball of mass $m$ connected with two ideal springs of force constant $k$ , kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration (in $m / s^{2}$ ) of ball?
supporting img

1

7 m / s^{2}

2

5 m / s^{2}

3

10 m / s^{2}

4

3 m / s^{2}

Explanation:

Let initial extension in spring is $x_{0}$ , then
$2 k x_{0} \cos 45^{\circ} = m g \sin 45^{\circ} \Rightarrow 2 k x_{0} = m g$
Let after cutting right spring, according of ball is $A_{1}$ along the inclined plane and $a_{2}$ perpendicular to $a_{1}$
$\Rightarrow \frac{m g}{\sqrt{2}} - \frac{m g}{2} \frac{1}{\sqrt{2}} = m a_{1}$
$\Rightarrow a_{1} = \frac{g}{2 \sqrt{2}}$
Then
$m g \sin 45^{\circ} \Rightarrow k x_{0} \cos 45^{\circ} = m a_{1}$
$k x_{0} \sin 45^{\circ} = m a_{2} \Rightarrow a_{2} = \frac{g}{2 \sqrt{2}}$
Net acceleration $a = \sqrt{a_{1}^{2} + a_{2}^{2}} = \frac{g}{2} = 5 m / s^{2}$

PHXI05:LAWS OF MOTION

363098 A block of mass 3 $k g$ is initially in equilibrium and is hanging by two identical springs $A$ and $B$ as shown in figure. If spring $A$ is cut from lower point at $t = 0$ then, find acceleration of block in $m s^{- 2}$ at $t = 0$ . (ignore rotational effect.)
supporting img

1

5

2

10

3

15

4

0

Explanation:

Before cutting the spring
$2 k x = 3 g \Rightarrow k x = 15$
After cutting
$a = \frac{3 g - k x}{3} = \frac{15}{3} = 5 m / s^{2}$
supporting img

PHXI05:LAWS OF MOTION

363099 Two masses $10 k g$ and $20 k g$ respectively are connected by a massless spring as shown in figure. A force of $200 N$ acts on the $20 k g$ mass. At the instant shown is figure, the $10 k g$ mass has acceleration of $12 m / s^{2} .$ The value of acceleration of $20 k g$ mass is
supporting img

1

4 m / s^{2}

2

10 m / s^{2}

3

20 m / s^{2}

4

30 m / s^{2}

Explanation:

Equation of motion of mass $m_{1} = 10 k g$ is
$F_{1} = m_{1} a_{1} = 10 \times 12 = 120 N$
Force on $10 k g$ mass is $120 N$ to the right. As
action and reaction are equal and opposite, the
reactions force $F$ - on $20 k g$ mass $F = 120 N$ to
the left.
$∴$ Equation of motion of mass
$m_{2} = 20 k g$ is
$200 - F = 20 a_{2}$
$200 - 120 = 20 a_{2}$
$20 a_{2} = 80$
$a_{2} = \frac{80}{20} = 4 m / s^{2}$